Transcript Section 2-2
5-Minute Check on Lesson 2-1b
1. Statistics are from samples
______ and parameters are from ________
populations
2. In a uniform distribution everything is equally
______ likely.
3. If a distribution is skewed right, which is greater, the mean or
the median and why?
mean. It is pulled toward the tail
(right and larger numbers)
1
4. The area under a density function is equal to ____
5. Name a common uniform probability example
a six-sided dice, coin (heads or tails)
6. Uniform probability distributions are of what types of
quantitative variables?
discrete and continuous!
Click the mouse button or press the Space Bar to display the answers.
Lesson 2 - 2
Normal Distributions
Objectives
DESCRIBE and APPLY the 68-95-99.7 Rule
DESCRIBE the standard Normal Distribution
PERFORM Normal distribution calculations
ASSESS Normality
Vocabulary
• 68-95-99.7 Rule (or Empirical Rule) – given a density curve is
normal (or population is normal), then the following is true:
within plus or minus one standard deviation is 68% of data
within plus or minus two standard deviation is 95% of data
within plus or minus three standard deviation is 99.7% of data
• Inverse Normal – calculator function that allows you to find a
data value given the area under the curve (percentage)
• Normal curve – special family of bell-shaped, symmetric
density curves that follow a complex formula
• Standard Normal Distribution – a normal distribution with a
mean of 0 and a standard deviation of 1
Normal Curves
• Two normal curves with different means (but
the same standard deviation) [on left]
– The curves are shifted left and right
• Two normal curves with different standard
deviations (but the same mean) [on right]
– The curves are shifted up and down
Normal Density Curve Properties
• It is symmetric about its mean, μ
• Because mean = median = mode, the highest point
occurs at x = μ
• It has inflection points at μ – σ and μ + σ
• Area under the curve = 1
• Area under the curve to the right of μ equals the area
under the curve to the left of μ, which equals ½
• As x increases or decreases without bound (gets
farther away from μ), the graph approaches, but
never reaches the horizontal axis (like approaching
an asymptote)
• The Empirical Rule (68-95-99.7) applies
Empirical Rule
μ ± 3σ
μ ± 2σ
μ±σ
99.7%
95%
68%
2.35%
0.15%
μ - 3σ
34%
13.5%
μ - 2σ
μ-σ
2.35%
34%
0.15%
13.5%
μ
μ+σ
μ + 2σ
μ + 3σ
Normal Probability Density Function
1
-(x – μ)2
y = -------- e 2σ2
√2π
where μ is the mean and σ is the standard deviation of the random variable x
Area under a Normal Curve
The area under the normal curve for any interval of
values of the random variable X represents either
• The proportion of the population with the
characteristic described by the interval of values or
• The probability that a randomly selected individual
from the population will have the characteristic
described by the interval of values
[the area under the curve is either a proportion or the
probability]
Standardizing a Normal Random Variable
Z statistic:
X-μ
Z = ----------σ
where μ is the mean and σ is the standard deviation of
the random variable X
Z is normally distributed with mean of 0 and standard
deviation of 1
Note: we are going to use tables (for Z statistics) not the
normal PDF!!
Or our calculator (see next chart)
Example 1
A random variable x is normally distributed with μ=10
and σ=3.
a. Compute Z for x1 = 8 and x2 = 12
8 – 10
-2
Z = ---------- = ----- = -0.67
3
3
12 – 10
2
Z = ----------- = ----- = 0.67
3
3
b. If the area under the curve between x1 and x2 is
0.495, what is the area between z1 and z2?
0.495
The Standard Normal Table
• Because all Normal distributions are the same when
we standardize, we can find areas under any Normal
curve from a single table.
Definition:
The Standard Normal Table
Table A is a table of areas under the standard Normal curve. The table
entry for each value z is the area under the curve to the left of z.
Suppose we want to find the
proportion of observations from the
standard Normal distribution that are
less than 0.81.
We can use Table A:
Z
.00
.01
.02
0.7
.7580
.7611
.7642
0.8
.7881
.7910
.7939
0.9
.8159
.8186
.8212
P(z < 0.81) = .7910
Normal Distributions on TI-83
• normalpdf pdf = Probability Density Function
This function returns the probability of a single value of
the random variable x. Use this to graph a normal curve.
Using this function returns the y-coordinates of the normal
curve.
• Syntax: normalpdf (x, mean, standard deviation)
taken from
http://mathbits.com/MathBits/TISection/Statistics2/normal
distribution.htm
• Remember the cataloghelp app on your calculator
– Hit the + key instead of enter when the item is highlighted
Normal Distributions on TI-83
• normalcdf cdf = Cumulative Distribution Function
This function returns the cumulative probability from zero
up to some input value of the random variable x.
Technically, it returns the percentage of area under a
continuous distribution curve from negative infinity to the
x. You can, however, set the lower bound.
• Syntax: normalcdf (lower bound, upper bound,
mean, standard deviation)
(note: lower bound is optional and we can use -E99
for negative infinity and E99 for positive infinity)
Normal Distributions on TI-83
• invNorm
inv = Inverse Normal PDF
This function returns the x-value given the probability
region to the left of the x-value. (0 < area < 1 must be
true.) The inverse normal probability distribution
function will find the precise value at a given percent based
upon the mean and standard deviation.
• Syntax: invNorm (probability, mean, standard
deviation)
Properties of the Standard Normal Curve
• It is symmetric about its mean, μ = 0, and has a
standard deviation of σ = 1
• Because mean = median = mode, the highest point
occurs at μ = 0
• It has inflection points at μ – σ = -1 and μ + σ = 1
• Area under the curve = 1
• Area under the curve to the right of μ = 0 equals the
area under the curve to the left of μ, which equals ½
• As Z increases the graph approaches, but never
reaches 0 (like approaching an asymptote). As Z
decreases the graph approaches, but never reaches, 0.
• The Empirical Rule (68-95-99.7) applies
Calculate the Area Under the Standard
Normal Curve
• Three different area calculations
– Find the area to the left of a value
– Find the area to the right of a value
– Find the area between two values
• There are several ways to calculate the area under the
standard normal curve
– What does not work – some kind of a simple formula
– We can use a table (such as Table IV on the inside back cover)
– We can use technology (a calculator or software)
• Using technology is preferred
Normal Distribution Calculations
How to Solve Problems Involving Normal Distributions
State: Express the problem in terms of the observed variable x.
Plan: Draw a picture of the distribution and shade the area of
interest under the curve.
Do: Perform calculations.
•Standardize x to restate the problem in terms of a standard
Normal variable z.
•Use Table A and the fact that the total area under the curve
is 1 to find the required area under the standard Normal curve.
Conclude: Write your conclusion in the context of the problem.
Obtaining Area under Standard Normal Curve
Approach
Graphically
Solution
Shade the area to the left of za
Use Table IV to find the row and
column that correspond to za. The
area is the value where the row
and column intersect.
Find the area to the
left of za
P(Z < a)
Normcdf(-E99,a,0,1)
a
Shade the area to the right of za
Find the area to the
right of za
Use Table IV to find the area to the
left of za. The area to the right of za
is 1 – area to the left of za.
Normcdf(a,E99,0,1) or
1 – Normcdf(-E99,a,0,1)
P(Z > a) or
1 – P(Z < a)
a
Shade the area between za and zb
Find the area
between za and zb
Use Table IV to find the area to the
left of za and to the left of za. The
area between is areazb – areaza.
Normcdf(a,b,0,1)
P(a < Z < b)
a
b
Example 2
Determine the area under the standard
normal curve that lies to the left of
a
a) Z = -3.49
Normalcdf(-E99,-3.49) = 0.000242
b) Z = -1.99
Normalcdf(-E99,-1.99) = 0.023295
c) Z = 0.92
Normalcdf(-E99,0.92) = 0.821214
d) Z = 2.90
Normalcdf(-E99,2.90) = 0.998134
Example 3
Determine the area under the standard
normal curve that lies to the right of
a) Z = -3.49
Normalcdf(-3.49,E99) = 0.999758
b) Z = -0.55
Normalcdf(-0.55,E99) = 0.70884
c) Z = 2.23
Normalcdf(2.23,E99) = 0.012874
d) Z = 3.45
Normalcdf(3.45,E99) = 0.00028
a
Example 4
Find the indicated probability of the
standard normal random variable Z
a
a) P(-2.55 < Z < 2.55)
Normalcdf(-2.55,2.55) = 0.98923
b) P(-0.55 < Z < 0)
Normalcdf(-0.55,0) = 0.20884
c) P(-1.04 < Z < 2.76)
Normalcdf(-1.04,2.76) = 0.84794
b
Example 5
Find the Z-score such that the area under the
standard normal curve to the left is 0.1.
invNorm(0.1) = -1.282 = a
a
Find the Z-score such that the area under the
standard normal curve to the right is 0.35.
invNorm(1-0.35) = 0.385
a
Summary and Homework
• Summary
– All normal distributions follow empirical rule
– Standard normal has mean = 0 and StDev = 1
– Table A gives you proportions that are less than z
• Homework
– Day 1: 41, 43, 45, 47, 49, 51
5-Minute Check on Lesson 2-2a
1. What is the mean and standard deviation of Z?
mean, = 0 and standard deviation, = 1
Given the following distributions: A~N(4,1), B~N(10,4) C~N(6,8)
2. Which is the tallest?
distribution A (it has smallest )
3. Which is the widest?
distribution C (it has largest )
68 , 95
99.7 rule.
4. The Empirical Rule is also known as the __
__ , ___
P( z > a) = 1 – P(z < a)
= 1 – 0.251
= 0.749
6. In distribution B, what is the area to the left of 10?
5. Given P(z < a) = 0.251, find P(z > a)
0.5
(half area is to left of mean)
Click the mouse button or press the Space Bar to display the answers.
Finding the Area under
any Normal Curve
• Draw a normal curve and shade the desired area
• Convert the values of X to Z-scores
using Z = (X – μ) / σ
• Draw a standard normal curve and shade the area
desired
• Find the area under the standard normal curve. This
area is equal to the area under the normal curve drawn
in Step 1
• Using your calculator, normcdf(-E99,x,μ,σ)
Given Probability Find the
Associated Random Variable Value
Procedure for Finding the Value of a Normal Random
Variable Corresponding to a Specified Proportion,
Probability or Percentile
• Draw a normal curve and shade the area
corresponding to the proportion, probability or
percentile
• Use Table IV to find the Z-score that corresponds to
the shaded area
• Obtain the normal value from the fact that X = μ + Zσ
• Using your calculator, invnorm(p(x),μ,σ)
Example 1
For a general random variable X with
μ=3
σ=2
a. Calculate Z
Z = (6-3)/2 = 1.5
b. Calculate P(X < 6)
so P(X < 6) = P(Z < 1.5) = 0.9332
Normcdf(-E99,6,3,2) or Normcdf(-E99,1.5)
Example 2
For a general random variable X with
μ = -2
σ=4
a. Calculate Z
Z = [-3 – (-2) ]/ 4 = -0.25
b. Calculate P(X > -3)
P(X > -3) = P(Z > -0.25) = 0.5987
Normcdf(-3,E99,-2,4)
Example 3
For a general random variable X with
– μ=6
– σ=4
calculate P(4 < X < 11)
P(4 < X < 11) = P(– 0.5 < Z < 1.25) = 0.5858
Converting to z is a waste of time for these
Normcdf(4,11,6,4)
Example 4
For a general random variable X with
– μ=3
– σ=2
find the value x such that P(X < x) = 0.3
x = μ + Zσ
Using the tables:
0.3 = P(Z < z) so z = -0.525
x = 3 + 2(-0.525)
so x = 1.95
invNorm(0.3,3,2) = 1.9512
Example 5
For a general random variable X with
– μ = –2
– σ=4
find the value x such that P(X > x) = 0.2
x = μ + Zσ
Using the tables:
P(Z>z) = 0.2 so P(Z<z) = 0.8
z = 0.842
x = -2 + 4(0.842)
so x = 1.368
invNorm(1-0.2,-2,4) = 1.3665
Example 6
For random variable X with
μ=6
σ=4
a
b
Find the values that contain 90% of the data
around μ
x = μ + Zσ
Using the tables: we know that z.05 = 1.645
x = 6 + 4(1.645)
so x = 12.58
x = 6 + 4(-1.645)
so x = -0.58
P(–0.58 < X < 12.58) = 0.90
invNorm(0.05,6,4) = -0.5794
invNorm(0.95,6,4) = 12.5794
Summary and Homework
• Summary
– Using a calculator we can avoid converting to zvalues before calculating the area under the
normal curves
– Calculator gives you proportions between any two
values (-e99 and e99 represent - and )
• Homework
– Day 2: 53, 55, 57, 59
5-Minute Check on Lesson 2-2b
Given a normal distribution with = 4 and = 2, Find
1. P(x < 2)
via TI: normalcdf(-e99,2,4,2) = 0.1587
2. P(x > 5)
via TI: normalcdf(5,e99,4,2) = 0.3085
3. P(1< x < 5)
via TI: normalcdf(1,5,4,2) = 0.6247
4. x, if P(x) = 0.95
via TI: invNorm(0.95,4,2) = 7.290
5. x, if P(x) = 0.05
via TI: invNorm(0.05,4,2) = 0.710
Click the mouse button or press the Space Bar to display the answers.
Is Data Normally Distributed?
• For small samples we can readily test it on our
calculators with Normal probability plots
• Large samples are better down using computer
software doing similar things
Normality Plots
• Most software packages can construct Normal
probability plots. These plots are constructed by
plotting each observation in a data set against its
corresponding percentile’s z-score.
Interpreting Normal Probability Plots
If the points on a Normal probability plot lie close to a straight line, the
plot indicates that the data are Normal. Systematic deviations from a
straight line indicate a non-Normal distribution. Outliers appear as points
that are far away from the overall pattern of the plot.
TI-83 Normality Plots
•
•
•
•
•
Enter raw data into L1
Press 2nd ‘Y=‘ to access STAT PLOTS
Select 1: Plot1
Turn Plot1 ON by highlighting ON and pressing ENTER
Highlight the last Type: graph (normality) and hit
ENTER. Data list should be L1 and the data axis should
be x-axis
• Press ZOOM and select 9: ZoomStat
Does it look pretty linear? (hold a piece of paper up to it)
Non-Normal Plots
• Both of these show that this particular data
set is far from having a normal distribution
– It is actually considerably skewed right
Example 1: Normal or Not?
Roughly Normal (linear in mid-range) with two possible
outliers on extremes
Example 2: Normal or Not?
Not Normal (skewed right); three possible outliers on
upper end
Example 3: Normal or Not?
Roughly Normal (very linear in mid-range)
Example 4: Normal or Not?
Roughly Normal (linear in mid-range) with deviations
on each extreme
Example 5: Normal or Not?
Not Normal (skewed right) with 3 possible outliers
Example 6: Normal or Not?
Roughly Normal (very linear in midrange) with 2
possible outliers
Summary and Homework
• Summary
– Calculator gives you proportions between any two
values (-e99 and e99 represent - and )
– Assess distribution’s potential normality by
• comparing with empirical rule
• normality probability plot (using calculator)
• Homework
– Day 3: 63, 65, 66, 68-74